字符串作为\\ x03 \\ x00 \\ x00 \\ x00到整数

Question

Helo:

I have a question, that I solved in c, but now I want to do it in hava: I have a String like: '\\x03\\x00\\x00\\x00' This is representing an hexadecimal value of a integer. I transform to 0x03\\0x00... And now I want to obtain the integer, but I don't know how to do it in java

could someone give me some idea ?

Thanks in advance

(Is it some way to use this format ('\\x03\\x00\\x00\\x00' ) directly without use byte[] arrays? and in C can I use this format directly to build a integer (int)?)

Answer 1

Here is my one line solution

Integer.parseInt("\\x03\\x00\\x00\\x00".replaceAll("\\\\x", ""), 16)

Of course, you might wrap this in a function.

Answer 2

I'd do it is something like this:

Scanner sc = new Scanner(inputString);
int value = 0;
for (int i = 0; i < 4; i++) {
    sc.next("\\\\x");
    byte b = sc.nextByte(16);
    value = (value << 8) | (b & 0xff);
}
if (sc.hasNext()) {
    throw new SomeException("Unexpected stuff at the end");
}

( UPDATE - corrected a sign extension bug ... and to detected unexpected stuff at the end of the input string.)

Now, if this encounters input that it can't cope with, it will throw an exception. In this case it is designed to match a sequence of EXACTLY 4 bytes.

But (IMO) that is preferable to accepting the bad input and producing an incorrect result. That is the risk with clever solutions that use pattern matching tricks to avoid doing proper parsing.

---

Thanks very much for the answers I don't understand this solution completely and it give me errors in sc.next("\\\\x");

The errors will be because you are using this on a string that has less than 4 bytes in it, or that has something that simply doesn't match the format of your example. If you want the code to handle sequences of less than (or more than) 4 bytes, you will need to change it.

I would like to know why you make: value=(value <<8) | b value=(value <<8) | b (but i'm working to understand it) thanks very much

Well there was a bug in that. I corrected it to

value = (value << 8) | (b & 0xff);

It is doing some bit-twiddling to tack the byte b onto the least significant end of the number in value . Read up on the Java << shift operator and the bit-wise & and | operators.